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Math Help - Integration help

  1. #1
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    Integration help

    So is it safe to say that we can use integration by substitution???
    Attached Thumbnails Attached Thumbnails Integration help-screen-shot-2012-12-13-8.09.48-pm.png  
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  2. #2
    MHF Contributor MarkFL's Avatar
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    Re: Integration help

    I would simplify the integrand by factoring \sqrt{t} from the numerator, then dividing out the common factors of the numerator and denominator. Or you might find it easier to just split the fraction, then reduce each term.
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  3. #3
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    Re: Integration help

    carry out the division
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  4. #4
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    Re: Integration help

    Can' i just put √t on the numerator and make it t^-1/2 and multiply (t + 2t^2)t^-1/2 together and integrate????
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  5. #5
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    Re: Integration help

    Can' i just put √t on the numerator and make it t^-1/2 and multiply (t + 2t^2)t^-1/2 together and integrate????.....
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  6. #6
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    Re: Integration help

    Yes, that's another way of splitting the fraction.
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  7. #7
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    Re: Integration help

    yup.

    you'd end up with \frac{t}{t^{1/2}}+\frac{2t^2}{t^{1/2}}=t^{1/2}+2*t^{3/2}

    or \int t^{1/2}+2*t^{3/2}dx
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  8. #8
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    Re: Integration help

    then im authorized to integrate after that step right?
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  9. #9
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    Re: Integration help

    Yes, now it is just a matter of applying the power rule term by term.
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  10. #10
    Senior Member x3bnm's Avatar
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    Re: Integration help

    Quote Originally Posted by asilvester635 View Post
    So is it safe to say that we can use integration by substitution???
    You can do this with the help of integration by parts.

    Now after simplification:

    \int \frac{t + 2t^2}{\sqrt{t}}\;\; dt = \int (\sqrt{t} + 2t\sqrt{t})\;\;dt = \int \sqrt{t}\;\;dt + \int  2t\sqrt{t}\;\;dt

    We divide this integration in two parts. First:

    \int \sqrt{t}\;\;dt =  \frac{2}{3}t^{\frac{3}{2}} + C_1

    Then for \int 2t\sqrt{t}\;\; dt we use integration by parts:

    \text{Let } u = t \;\;\therefore du = dt

    \text{Let } dv = 2\sqrt{t}\;\; dt\;\; \therefore v = \frac{4}{3}t^{\frac{3}{2}}

    Now:

    \begin{align*}\int  2t\sqrt{t}\;\;dt =& \frac{4}{3}t^{\frac{5}{2}} - \int \frac{4}{3}t^{\frac{3}{2}}\;\; dt \\ =& \frac{4}{3}t^{\frac{5}{2}} - \frac{8}{15}t^{\frac{5}{2}} + C_2 \\ =& \frac{4}{5}t^{\frac{5}{2}}+ C_2...........\text{[After simplicfication]}\end{align*}

    So:

    \therefore \int \frac{t + 2t^2}{\sqrt{t}}\;\; dt =  \frac{2}{3}t^{\frac{3}{2}} + \frac{4}{5}t^{\frac{5}{2}} + C


    Check:

    integration (t + 2*t^2)/(t)^(1/2) dt - Wolfram|Alpha
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